Problem: $ 9^{-\frac{3}{2}}$
$= \left(\dfrac{1}{9}\right)^{\frac{3}{2}}$ $= \left(\left(\dfrac{1}{9}\right)^{\frac{1}{2}}\right)^{3}$ To simplify $\left(\dfrac{1}{9}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left(? \right)^{2}=\dfrac{1}{9}$ To simplify $\left(\dfrac{1}{9}\right)^{\frac{1}{2}}$ , figure out what goes in the blank: $\left({\dfrac{1}{3}}\right)^{2}=\dfrac{1}{9}$ so $ \left(\dfrac{1}{9}\right)^{\frac{1}{2}}=\dfrac{1}{3}$ So $\left(\dfrac{1}{9}\right)^{\frac{3}{2}}=\left(\left(\dfrac{1}{9}\right)^{\frac{1}{2}}\right)^{3}=\left(\dfrac{1}{3}\right)^{3}$ $= \left(\dfrac{1}{3}\right)^{3}$ $= \left(\dfrac{1}{3}\right)\cdot\left(\dfrac{1}{3}\right)\cdot \left(\dfrac{1}{3}\right)$ $= \dfrac{1}{9}\cdot\left(\dfrac{1}{3}\right)$ $= \dfrac{1}{27}$